A gambling game works as follows: you flip a fair coin and roll a fair six-sided die. You will be paid $2 if you roll a 6 and $1 if you get a head and an odd number of spots. Otherwise, you have to pay $2.50.
a) What is the probability of rolling a die greater than 1?
b) What is the probability of winning money from this game? Show all your work.
c) How much can you expect to win if you played this game 20 times?
12 outcomes ... H or T, and 1 thru 6
2 outcomes pay $2
... P = 2/12 = 1/6
... expected return ... 2/6 or $1/3
3 outcomes pay $1
... P = 3/12 = 1/4
... expected return ... $1/4
To answer these questions, we need to first find the probabilities involved in the different events.
a) What is the probability of rolling a die greater than 1?
A six-sided die has numbers from 1 to 6. So, to find the probability of rolling a die greater than 1, we need to count the number of favorable outcomes (i.e., numbers greater than 1) and divide it by the total number of possible outcomes.
Favorable outcomes: {2, 3, 4, 5, 6} (5 numbers)
Total outcomes: {1, 2, 3, 4, 5, 6} (6 numbers)
Therefore, the probability of rolling a die greater than 1 is 5/6.
b) What is the probability of winning money from this game? Show all your work.
To win money, either we need to roll a 6 or get a head and an odd number of spots. Let's calculate the probability of each individual event and sum them up.
Event 1: Rolling a 6
The probability of rolling a 6 on a fair six-sided die is 1/6.
Event 2: Getting a head and an odd number of spots
The probability of getting a head is 1/2 (since a fair coin has 2 sides).
The probability of getting an odd number on the die is 3/6 (since there are 3 odd numbers on a six-sided die).
To find the probability of both events happening together (i.e., getting a head and an odd number), we multiply the probabilities of each event:
Probability of getting a head and an odd number = (1/2) * (3/6) = 1/4
Adding the probabilities of Event 1 and Event 2, we get:
Probability of winning money = 1/6 + 1/4 = 5/12
Therefore, the probability of winning money from this game is 5/12.
c) How much can you expect to win if you played this game 20 times?
To calculate the expected winnings, we need to multiply the probability of each outcome by its respective amount.
Outcome 1: Winning $2 (rolling a 6)
Probability = 1/6
Winnings = $2
Outcome 2: Winning $1 (getting a head and an odd number)
Probability = 1/4
Winnings = $1
Outcome 3: Losing $2.50
Probability = 1 - (Probability of winning money) = 1 - (5/12) = 7/12 (probability of losing money)
Loss = -$2.50
Expected winnings per game = (Probability of Outcome 1 * Winnings from Outcome 1) + (Probability of Outcome 2 * Winnings from Outcome 2) + (Probability of Outcome 3 * Loss from Outcome 3)
Expected winnings per game = (1/6 * $2) + (1/4 * $1) + (7/12 * -$2.50)
Now, we can calculate the expected winnings in 20 games by multiplying the expected winnings per game by the number of games played:
Expected winnings in 20 games = 20 * ((1/6 * $2) + (1/4 * $1) + (7/12 * -$2.50))
This will give us the expected amount you can win or lose if you played this game 20 times.